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3 years ago

8

Find the length indicated. find CE

1 answer:

torisob [31]3 years ago

6 0

Okay, to find length CE, your going to know the value of <em>x</em>. Length BC + CE = BD + DE.
3x+47+x+26=27+x+10
Simplify the equation to get
4x+73=37+x
you can choose one of four ways to continue, but I will choose to subtract x
3x+73=37
Subtract 73 from both sides of the equal sign
3x=-36
divide by 3 on both sides of the equal sign to get the value of x
x=-12

Now, plug in -12 for x in length CE to get -12+26=14

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NEED HELP ASAP!!!

sergiy2304 [10]

14. To get the length of each side, you will have to find the square root of 189 to find what two numbers multiplied by itself will get to that number. The square root of 189 is about 13.7, to the nearest tenths is 14

7 0

3 years ago

What is the average rate of change of the function f(x)=5(2)^x from x = 1 to x = 5?

Mila [183]

Average rate of change refers to the slope of a function/equation.
To solve this, plug in for both x values and calculate the average, or mean.

x = 1
f(1) = 5(2)^1
= 5(2)
= 10

x = 5
f(5) = 5(2)^5
= 5(32)
= 160

Now, you calculate the average:
10 + 160 = 170
Divide by 2 and you get 85.

5 0

3 years ago

For what value(s) of k is the linear system consistent? (Enter your answers as a comma-separated list.)

AleksAgata [21]

Answer:

Step-by-step explanation:

Given

$6x_1-9x_2=8$

$9x_1+kx_2=-1$

The given system is $AX=B$ can be represented by

$\begin{bmatrix}6 &-9 \\ 9 & k\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}8\\ -1\end{bmatrix}$

The given system is consistent when determinant of A is not equal to zero

$|A|$

$|A|=6k-(-81)=6k+81$

$k\neq \frac{-27}{2}$

i.e. system is consistent for all value of k except $k=\frac{-27}{2}$

$R-\frac{-27}{2}$

4 0

3 years ago

Read 2 more answers

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago

Write 7/50 as a decimal

Sophie [7]

0.14

You divide 7 by 50.

5 0

3 years ago

Read 2 more answers